SOLUTION: 2. Assume the body temperatures of healthy adults are normally distributed with a mean of 98.20 °F and a standard deviation of 0.62 °F (based on data from the University of Marylan

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Question 314444: 2. Assume the body temperatures of healthy adults are normally distributed with a mean of 98.20 °F and a standard deviation of 0.62 °F (based on data from the University of Maryland researchers).
a. If you have a body temperature of 99.00 °F, what is your percentile score?
b. Convert 99.00 °F to a standard score (or a z-score).
c. Is a body temperature of 99.00 °F unusual? Why or why not?
d. Fifty adults are randomly selected. What is the likelihood that the mean of their body temperatures is 97.98 °F or lower?
e. A person’s body temperature is found to be 101.00 °F. Is the result unusual? Why or why not? What should you conclude?
f. What body temperature is the 95th percentile?
g. What body temperature is the 5th percentile?
h. Bellevue Hospital in New York City uses 100.6 °F as the lowest temperature considered to indicate a fever. What percentage of normal and healthy adults would be considered to have a fever? Does this percentage suggest that a cutoff of 100.6 °F is appropriate?


Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
a.mu=98.20
sigma=0.62
x=99
z=%28x-mu%29%2Fsigma=%2899-98.2%29%2F0.62=1.33
P%281.33%29=0.9087
91st percentile.
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b. z=1.33
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c. Depends on what you consider normal, unusual, etc., you should make that determination and be able to defend your position.
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d. Set up a t test.
The null hypothesis is that the means are the same.
N=50
X=97.98
S=sigma%2Fsqrt%28N-1%29=0.62%2Fsqrt%2850-1%29=0.62%2F7=0.00886
t=%28X-mu%29%2FS=%2897.98-98.2%29%2F0.00886=-2.484
Now the challenge is in working out the critical t value.
t(0.05,49)=1.647
t(0.01,49)=2.405
Depending on the critical t value you choose, you could reject the null hypothesis, in which case, you would say the means are not the same, there is a statistically significant difference in the findings.
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e. x=101
z=%28x-mu%29%2Fsigma=%28101-98.2%29%2F0.62=4.52
P%284.52%29=0.999998
Yes, that would be odd because very few people would normally have that temperature.
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f. P%28z%29=0.95
z=1.645=%28x-98.2%29%2F0.62
x=99.2
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g. What body temperature is the 5th percentile?
P%28z%29=0.05
z=-1.645=%28x-98.2%29%2F0.62
x=97.2
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h. z=%28x-mu%29%2Fsigma=%28100.6-98.2%29%2F0.62=1.33
P%28z%29=0.999968
Normally and healthy would be ones who have higher than 100.6 but are not sick.
P=1-0.999968=0.000032
That's pretty low, so you would think it's pretty unlikely that someone coming in off the street would have a 100.6 temperature and be normal. The cutoff is probably appropriate. Again check the caveats in section c.